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When $f\left( x\right) =\log x$ and section $\left[ 1,2\right]$ $$\lim _{n\rightarrow \infty }n\left[ \int _{a}^{b}f\left( x\right) dx-\dfrac{b-a}{n} \left\{\dfrac{f\left( a\right) +f\left( b\right) }{2}+\sum ^{n-1}_{k=1}f\left( a+\dfrac{\left( b-a\right) k}{n}\right) \right\} \right] =0$$ Use this to find the value of the limit: $$\lim _{n\rightarrow \infty }\dfrac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !}-①$$

I have assigned $f(x)$ and section $\left[ 1,2\right]$. $$\lim _{n\rightarrow \infty }n\left[ \dfrac{1}{2}\log 2-\dfrac{1}{n}\left\{\dfrac{1}{2}\log 2+\sum ^{n-1}_{k=1}\log \left( 1+\dfrac{k}{n}\right) \right\} \right]-②$$ I can't find common points of $①$ and $②$. So I don't know what to do next.

StubbornAtom
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langhtorn
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  • Are you sure of the $n$ factor at the start of the limit in front of the left bracket? It should be $\lim {\rightarrow \infty }\left[ \int _{a}^{b}f\left( x\right) dx-\dfrac{b-a}{n} \left{\dfrac{f\left( a\right) +f\left( b\right) }{2}+\sum ^{n-1}{k=1}f\left( a+\dfrac{\left( b-a\right) k}{n}\right) \right} \right] =0$ – mathcounterexamples.net Jul 29 '20 at 09:50
  • @AniruddhaDeb I want to solve $\lim _{n\rightarrow \infty }\dfrac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !}-①$. – langhtorn Jul 29 '20 at 09:56
  • @langhtorn I edited the question for more clarity now. – Aniruddha Deb Jul 29 '20 at 09:57
  • See related https://math.stackexchange.com/a/2847768/72031 The answer to your question should be $1/\sqrt{2}$. – Paramanand Singh Jul 29 '20 at 12:43

2 Answers2

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This answer is incorrect.

I'm leaving it here (instead of deleting it, got the advice here) just so that other solvers don't make the same mistake.

As rightly pointed out in the comments, This lemma should be used to obtain the correct answer of $1/\sqrt 2$

$$\begin{gather} L = \lim_{n \to \infty} \left( \frac{4}{e} \right)^n \cdot \left( \frac{n}{n+1} \cdot \frac{n}{n+2} ... \frac{n}{n+n}\right) \\ \log L = \lim_{n \to \infty} n \log \left( \frac 4e \right) + \sum_{r=1}^n\log\left(\frac{n}{n+r}\right) \\ \log L = \lim_{n \to \infty} n \left( \log \left( \frac 4e \right) - \frac 1n\sum_{r=1}^{n}\log\left(1+\frac rn\right)\right) \end{gather}$$ the term on the right is a Riemann sum, and it can be derived by using the trapezoidal rule to get the area for infinitesimal parts and summing them up. I'll leave this part up to you but the solution of the riemann sum is $\log\left( \frac 4e\right)$. This makes the limit $$\log L = \lim_{n \to \infty} n \left( \log \left( \frac 4e \right) - \log \left( \frac 4e \right) \right) \\ \log L = 0 \\ \boxed{L = 1}$$

Aniruddha Deb
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  • The ultimate answer may be correct (I'm not sure if it is or isn't), but the logic of $\lim_{n\to\infty}n(c-f(n))=\lim_{n\to\infty}n(c-\lim_{n\to\infty}f(n))=\lim_{n\to\infty}n(c-c)=0$ is not. – Barry Cipra Jul 29 '20 at 11:22
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    On further reflection, the correct limit is $L=1/\sqrt2$, so this is definitely a case where partially "distributing" the limit leads one astray. – Barry Cipra Jul 29 '20 at 11:38
  • You have made a typical mistake at the end. When you have a product of two terms where one tends to $\infty $ and other tends to $0$ then one can't decide the limit without further deeper analysis. See https://math.stackexchange.com/a/2847768/72031 for the right approach. – Paramanand Singh Jul 29 '20 at 12:45
  • Edited to say that it is incorrect. – Aniruddha Deb Jul 29 '20 at 13:17
  • Thank you. I can understand. – langhtorn Jul 30 '20 at 02:42
  • @langhtorn I would recommend you accept a different answer as the correct one: this one is clearly incorrect. – Aniruddha Deb Jul 30 '20 at 04:12
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$$a_n=\dfrac{2^{2n}e^{-n}n^{n}n!}{\left( 2n\right) !}\implies \log(a_n)=2n \log(2)-n+n \log(n)+\log(n!)-\log((2n)!)$$

Use Stirling approximation twice and continue with Taylor expansion. Yous should obtain $$\log(a_n)=-\frac{\log (2)}{2}+\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$ Just continue $$a_n=e^{\log(a_n)}=\frac{1}{\sqrt{2}}\left(1+\frac{1}{24 n}+\frac{1}{1152 n^2}\right)+O\left(\frac{1}{n^3}\right)$$

Claude Leibovici
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